The value of α, for which the equation x2-( sin α-2) x-(1+ sin α)=0 has roots whose sum of square is least, is

Question

The value of α, for which the equation x2-( sin α-2) x-(1+ sin α)=0 has roots whose sum of square is least, is (A)  (π /3)  (B)  (π /4)  (C)  (π /2)  (D)  (π /6)

Tardigrade Admin · Accepted Answer

Let the roots of the equation be p, q. Let S=p2+q2 =(p+q)2-2 p q ldots (i) Given equation is x2-( sin α-2) x-(1+ sin α)=0 ∴ p+q=( sin α-2), p q=-(1+ sin α) From Eq. (i), S=( sin α-2)2+2(1+ sin α) = sin 2 α-4 sin α+4+2+2 sin α = sin 2 α-2 sin α+6 ⇒ S=( sin α-1)2+5 This is least when sin α-1=0 ⇒ α=(π/2)

Q. The value of $α$ , for which the equation $x^{2} - (sin α - 2) x - (1 + sin α) = 0$ has roots whose sum of square is least, is

$\frac{π}{3}$

$\frac{π}{4}$

$\frac{π}{2}$

$\frac{π}{6}$

Q. The value of α, for which the equation x2−(sinα−2)x−(1+sinα)=0 has roots whose sum of square is least, is

3π​

4π​

2π​

6π​

Q. The value of $α$ , for which the equation $x^{2} - (sin α - 2) x - (1 + sin α) = 0$ has roots whose sum of square is least, is

$\frac{π}{3}$

$\frac{π}{4}$

$\frac{π}{2}$

$\frac{π}{6}$